Having a nice hot shower at the Off Grid Camper
Posted: Fri May 10, 2013 7:05 pm
I don't know everything about Troy's camper, but if it's a normal camper, it's got (or had) a 6 gallon hot water heater. I have had 4 RV's with such water heaters.
I've been wondering if Troy is overlooking a free (energy cost wise) hot shower.
My current camper for work, I have installed an electric probe heater in the drain plug. This is a product as I remember marketed under the name Lightening Rod. When I purchased it, they were available in 2 sizes. I got the smaller one intended for smaller tanks. I've noticed, even though I only got the 600 watt unit, it works just fine and it does not seem to take very long at all. I chose the smaller unit because I use a Honda 2000 generator and I didn't want to have to shut down the water heater to run the microwave or to let the refrigerator run on AC from the generator. Seemed like the smart thing to do at the time and I've been happy with that.
I have observed in a number of video's Troy mentions the nice charge controller from Arizona Wind and Sun has gone into "Absorption" charge mode fairly early in the day. Once the batteries are full, the rest of the energy above his needs is lost since he can store no more. In other words, this excess energy is not supplying any services, it's totally wasted.
Hey, sweep up all those messy electrons you're littering the land with!
So, I got to wondering:
How much energy is required to run the 600 watt heating probe?
How long would it have to run?
How many amps will this pull out of the battery bank if any?
How long will it take for the battery bank to recover?
So, lets get to calculating.
A single BTU is what is required to heat one pound of water and cause it to rise 1 degree Fahrenheit.
Water weighs about 8.34 pounds per gallon.
The 6 gallon heater contains 50 pounds of water (maybe a little less as some of it may be expansion chamber volume)
If we start at 70 degree and want to raise it 40 degrees to achieve 110 the math is like so:
50 pounds of water X 40 degrees temp rise = 2000 BTU's required.
A google search shows:
There are 3412 BTU's per 1000 watt hours or 1KWh.
2000 BTU Required /3412 Per KWH = 58% of 1000 watt hour or 580 watt hours. Or, one hour with a 580 watt heater.
Our 600 watt hour (600wh or .6kwh) heating element should increase the temperature of our 6 gallons of water in the tank 40 degrees in just short of 1 hour.
If we only shower on a sunny day, and we only shower after the charger reaches absorption mode, we can use the entire wattage of the solar panels to heat the water.
How many watts does Troy have? I'm not sure. But, lets say he has 400 watts for our calculation.
The difference between 600 watt demand and our 400 watt available supply is 200 watts. 200 watts at 12v is 200/12 = 16 amps.
Now in reality, he may have a few more watts and I'm not accounting for losses in the wiring and inverter. But suffice it to say, I am very confident for the 58 minutes Troy would have to turn on the 600 watt heating probe in the water heater, he'd be pulling no more than 20 amps out of his battery bank. 20 amps at 12v over one hour would be 20x12 or 240 watts net battery charge loss heating the water. 240 watts is .24kwh of energy.
If he has 400 watts going in from solar, this would mean when he turns off the heating probe, the charge would be returned to the battery bank in 240/400 of an hour or about
If his battery bank is 400 amps (Iforget it's size also). 400 amps x 12v = 4800 watts or 4.8kwh of capacity.
Using .24kwh of energy above production of power would drian his battery bank 5% (.24khw / 4.8kwh = .05 or 5%
Time to recharge after turning off the heating element is 240 watts lost / 400 watts input = 60% of an hour or 36 minutes.
Troy, all you need is a 600 watt lightening rod and you could be taking hot showers. No need to scream in that ice cold creek any more.
I've been wondering if Troy is overlooking a free (energy cost wise) hot shower.
My current camper for work, I have installed an electric probe heater in the drain plug. This is a product as I remember marketed under the name Lightening Rod. When I purchased it, they were available in 2 sizes. I got the smaller one intended for smaller tanks. I've noticed, even though I only got the 600 watt unit, it works just fine and it does not seem to take very long at all. I chose the smaller unit because I use a Honda 2000 generator and I didn't want to have to shut down the water heater to run the microwave or to let the refrigerator run on AC from the generator. Seemed like the smart thing to do at the time and I've been happy with that.
I have observed in a number of video's Troy mentions the nice charge controller from Arizona Wind and Sun has gone into "Absorption" charge mode fairly early in the day. Once the batteries are full, the rest of the energy above his needs is lost since he can store no more. In other words, this excess energy is not supplying any services, it's totally wasted.
Hey, sweep up all those messy electrons you're littering the land with!
So, I got to wondering:
How much energy is required to run the 600 watt heating probe?
How long would it have to run?
How many amps will this pull out of the battery bank if any?
How long will it take for the battery bank to recover?
So, lets get to calculating.
A single BTU is what is required to heat one pound of water and cause it to rise 1 degree Fahrenheit.
Water weighs about 8.34 pounds per gallon.
The 6 gallon heater contains 50 pounds of water (maybe a little less as some of it may be expansion chamber volume)
If we start at 70 degree and want to raise it 40 degrees to achieve 110 the math is like so:
50 pounds of water X 40 degrees temp rise = 2000 BTU's required.
A google search shows:
There are 3412 BTU's per 1000 watt hours or 1KWh.
2000 BTU Required /3412 Per KWH = 58% of 1000 watt hour or 580 watt hours. Or, one hour with a 580 watt heater.
Our 600 watt hour (600wh or .6kwh) heating element should increase the temperature of our 6 gallons of water in the tank 40 degrees in just short of 1 hour.
If we only shower on a sunny day, and we only shower after the charger reaches absorption mode, we can use the entire wattage of the solar panels to heat the water.
How many watts does Troy have? I'm not sure. But, lets say he has 400 watts for our calculation.
The difference between 600 watt demand and our 400 watt available supply is 200 watts. 200 watts at 12v is 200/12 = 16 amps.
Now in reality, he may have a few more watts and I'm not accounting for losses in the wiring and inverter. But suffice it to say, I am very confident for the 58 minutes Troy would have to turn on the 600 watt heating probe in the water heater, he'd be pulling no more than 20 amps out of his battery bank. 20 amps at 12v over one hour would be 20x12 or 240 watts net battery charge loss heating the water. 240 watts is .24kwh of energy.
If he has 400 watts going in from solar, this would mean when he turns off the heating probe, the charge would be returned to the battery bank in 240/400 of an hour or about
If his battery bank is 400 amps (Iforget it's size also). 400 amps x 12v = 4800 watts or 4.8kwh of capacity.
Using .24kwh of energy above production of power would drian his battery bank 5% (.24khw / 4.8kwh = .05 or 5%
Time to recharge after turning off the heating element is 240 watts lost / 400 watts input = 60% of an hour or 36 minutes.
Troy, all you need is a 600 watt lightening rod and you could be taking hot showers. No need to scream in that ice cold creek any more.